Six employees – A, B, C, D, E and F – of a firm entered an empty lift together at the ground floor (floor 0) of their office building at exactly 9:00:00 a.m. The building has floors 1 to 7 above the ground floor. Each employee got off at the floor of his or her own office, and no one else entered the lift during this trip. The lift moved only upwards, and stopped only at floors where at least one of the six got off.
The lift takes 5 seconds to travel from one floor to the next, and every stop lasts exactly 12 seconds. Employees get off the moment the lift arrives at a stop.
The following is also known:
1. B got off before every other employee, and no one got off with B.
2. After the lift's second stop, exactly three employees were left in the lift.
3. Exactly two employees got off at the lift's final (highest) stop.
4. A got off at an odd-numbered floor, along with exactly one other employee.
5. C and F got off at the same floor.
6. D's office is exactly two floors above A's office.
7. The sum of the numbers of the floors at which the lift stopped was 17.
The lift takes 5 seconds to travel from one floor to the next, and every stop lasts exactly 12 seconds. Employees get off the moment the lift arrives at a stop.
The following is also known:
1. B got off before every other employee, and no one got off with B.
2. After the lift's second stop, exactly three employees were left in the lift.
3. Exactly two employees got off at the lift's final (highest) stop.
4. A got off at an odd-numbered floor, along with exactly one other employee.
5. C and F got off at the same floor.
6. D's office is exactly two floors above A's office.
7. The sum of the numbers of the floors at which the lift stopped was 17.
Question 1
Single correct option
Who got off the lift along with A?
- A. C
- B. D
- C. E
- D. F
Show answer and solution
Answer: (C) E
Deduction: B leaves alone at stop 1 (clue 1). Clue 2 means three people have left after stop 2, so exactly two leave at stop 2. Of the remaining three, two leave at the top stop (clue 3), so one leaves at a third stop: 4 stops in all, with group sizes 1, 2, 1, 2.
The two pairs are {C, F} (clue 5) and A with one other (clue 4). A's partner cannot be D (clue 6), B, C or F, so it is E: pairs {A, E} and {C, F}; D is the single at stop 3. Since D is above A, {A, E} is stop 2 and {C, F} is the top stop.
Floors: b < a < a + 2 < c ≤ 7 with a odd and b + a + (a + 2) + c = 17. a = 1 leaves no room for b; a = 5 makes D = 7 with no room for c. So a = 3, D = 5, and b + c = 9 with c ∈ {6, 7}, b ≤ 2 → b = 2, c = 7.
| Stop | Floor | Got off |
|---|---|---|
| 1 | 2 | B |
| 2 | 3 | A, E |
| 3 | 5 | D |
| 4 | 7 | C, F |
A's companion is E.
The two pairs are {C, F} (clue 5) and A with one other (clue 4). A's partner cannot be D (clue 6), B, C or F, so it is E: pairs {A, E} and {C, F}; D is the single at stop 3. Since D is above A, {A, E} is stop 2 and {C, F} is the top stop.
Floors: b < a < a + 2 < c ≤ 7 with a odd and b + a + (a + 2) + c = 17. a = 1 leaves no room for b; a = 5 makes D = 7 with no room for c. So a = 3, D = 5, and b + c = 9 with c ∈ {6, 7}, b ≤ 2 → b = 2, c = 7.
| Stop | Floor | Got off |
|---|---|---|
| 1 | 2 | B |
| 2 | 3 | A, E |
| 3 | 5 | D |
| 4 | 7 | C, F |
A's companion is E.
Question 2
Numerical answer
How many seconds after 9:00:00 a.m. did D get off the lift?
Show answer and solution
Answer: 49
Deduction: B leaves alone at stop 1 (clue 1). Clue 2 means three people have left after stop 2, so exactly two leave at stop 2. Of the remaining three, two leave at the top stop (clue 3), so one leaves at a third stop: 4 stops in all, with group sizes 1, 2, 1, 2.
The two pairs are {C, F} (clue 5) and A with one other (clue 4). A's partner cannot be D (clue 6), B, C or F, so it is E: pairs {A, E} and {C, F}; D is the single at stop 3. Since D is above A, {A, E} is stop 2 and {C, F} is the top stop.
Floors: b < a < a + 2 < c ≤ 7 with a odd and b + a + (a + 2) + c = 17. a = 1 leaves no room for b; a = 5 makes D = 7 with no room for c. So a = 3, D = 5, and b + c = 9 with c ∈ {6, 7}, b ≤ 2 → b = 2, c = 7.
| Stop | Floor | Got off |
|---|---|---|
| 1 | 2 | B |
| 2 | 3 | A, E |
| 3 | 5 | D |
| 4 | 7 | C, F |
Timing: 0 → 2: 10 s (arrive 10 s), halt till 22 s; 2 → 3: 5 s (arrive 27 s), halt till 39 s; 3 → 5: 10 s → arrive at 49 s, when D gets off.
Tip: time of arrival at a floor = 5 × (floor number) + 12 × (number of earlier stops) = 25 + 24 = 49.
The two pairs are {C, F} (clue 5) and A with one other (clue 4). A's partner cannot be D (clue 6), B, C or F, so it is E: pairs {A, E} and {C, F}; D is the single at stop 3. Since D is above A, {A, E} is stop 2 and {C, F} is the top stop.
Floors: b < a < a + 2 < c ≤ 7 with a odd and b + a + (a + 2) + c = 17. a = 1 leaves no room for b; a = 5 makes D = 7 with no room for c. So a = 3, D = 5, and b + c = 9 with c ∈ {6, 7}, b ≤ 2 → b = 2, c = 7.
| Stop | Floor | Got off |
|---|---|---|
| 1 | 2 | B |
| 2 | 3 | A, E |
| 3 | 5 | D |
| 4 | 7 | C, F |
Timing: 0 → 2: 10 s (arrive 10 s), halt till 22 s; 2 → 3: 5 s (arrive 27 s), halt till 39 s; 3 → 5: 10 s → arrive at 49 s, when D gets off.
Tip: time of arrival at a floor = 5 × (floor number) + 12 × (number of earlier stops) = 25 + 24 = 49.
Question 3
Single correct option
At how many of the floors 1 to 7 did the lift pass without stopping?
- A. 2
- B. 3
- C. 4
- D. 5
Show answer and solution
Answer: (B) 3
Deduction: B leaves alone at stop 1 (clue 1). Clue 2 means three people have left after stop 2, so exactly two leave at stop 2. Of the remaining three, two leave at the top stop (clue 3), so one leaves at a third stop: 4 stops in all, with group sizes 1, 2, 1, 2.
The two pairs are {C, F} (clue 5) and A with one other (clue 4). A's partner cannot be D (clue 6), B, C or F, so it is E: pairs {A, E} and {C, F}; D is the single at stop 3. Since D is above A, {A, E} is stop 2 and {C, F} is the top stop.
Floors: b < a < a + 2 < c ≤ 7 with a odd and b + a + (a + 2) + c = 17. a = 1 leaves no room for b; a = 5 makes D = 7 with no room for c. So a = 3, D = 5, and b + c = 9 with c ∈ {6, 7}, b ≤ 2 → b = 2, c = 7.
| Stop | Floor | Got off |
|---|---|---|
| 1 | 2 | B |
| 2 | 3 | A, E |
| 3 | 5 | D |
| 4 | 7 | C, F |
The lift stopped at floors 2, 3, 5 and 7, so it passed floors 1, 4 and 6 without stopping: 3 floors.
The two pairs are {C, F} (clue 5) and A with one other (clue 4). A's partner cannot be D (clue 6), B, C or F, so it is E: pairs {A, E} and {C, F}; D is the single at stop 3. Since D is above A, {A, E} is stop 2 and {C, F} is the top stop.
Floors: b < a < a + 2 < c ≤ 7 with a odd and b + a + (a + 2) + c = 17. a = 1 leaves no room for b; a = 5 makes D = 7 with no room for c. So a = 3, D = 5, and b + c = 9 with c ∈ {6, 7}, b ≤ 2 → b = 2, c = 7.
| Stop | Floor | Got off |
|---|---|---|
| 1 | 2 | B |
| 2 | 3 | A, E |
| 3 | 5 | D |
| 4 | 7 | C, F |
The lift stopped at floors 2, 3, 5 and 7, so it passed floors 1, 4 and 6 without stopping: 3 floors.
Question 4
Single correct option
What is the sum of the floor numbers of the offices of C, D and E?
- A. 13
- B. 15
- C. 16
- D. 17
Show answer and solution
Answer: (B) 15
Deduction: B leaves alone at stop 1 (clue 1). Clue 2 means three people have left after stop 2, so exactly two leave at stop 2. Of the remaining three, two leave at the top stop (clue 3), so one leaves at a third stop: 4 stops in all, with group sizes 1, 2, 1, 2.
The two pairs are {C, F} (clue 5) and A with one other (clue 4). A's partner cannot be D (clue 6), B, C or F, so it is E: pairs {A, E} and {C, F}; D is the single at stop 3. Since D is above A, {A, E} is stop 2 and {C, F} is the top stop.
Floors: b < a < a + 2 < c ≤ 7 with a odd and b + a + (a + 2) + c = 17. a = 1 leaves no room for b; a = 5 makes D = 7 with no room for c. So a = 3, D = 5, and b + c = 9 with c ∈ {6, 7}, b ≤ 2 → b = 2, c = 7.
| Stop | Floor | Got off |
|---|---|---|
| 1 | 2 | B |
| 2 | 3 | A, E |
| 3 | 5 | D |
| 4 | 7 | C, F |
C = 7, D = 5, E = 3 → sum = 15.
Trap: 17 is the sum of the stop floors (2 + 3 + 5 + 7), which includes B's floor and not E's.
The two pairs are {C, F} (clue 5) and A with one other (clue 4). A's partner cannot be D (clue 6), B, C or F, so it is E: pairs {A, E} and {C, F}; D is the single at stop 3. Since D is above A, {A, E} is stop 2 and {C, F} is the top stop.
Floors: b < a < a + 2 < c ≤ 7 with a odd and b + a + (a + 2) + c = 17. a = 1 leaves no room for b; a = 5 makes D = 7 with no room for c. So a = 3, D = 5, and b + c = 9 with c ∈ {6, 7}, b ≤ 2 → b = 2, c = 7.
| Stop | Floor | Got off |
|---|---|---|
| 1 | 2 | B |
| 2 | 3 | A, E |
| 3 | 5 | D |
| 4 | 7 | C, F |
C = 7, D = 5, E = 3 → sum = 15.
Trap: 17 is the sum of the stop floors (2 + 3 + 5 + 7), which includes B's floor and not E's.
Question 5
Numerical answer
After the halt at its final stop of this trip, the lift returned directly to the ground floor without stopping. How many seconds after 9:00:00 a.m. did it reach the ground floor?
Show answer and solution
Answer: 118
Deduction: B leaves alone at stop 1 (clue 1). Clue 2 means three people have left after stop 2, so exactly two leave at stop 2. Of the remaining three, two leave at the top stop (clue 3), so one leaves at a third stop: 4 stops in all, with group sizes 1, 2, 1, 2.
The two pairs are {C, F} (clue 5) and A with one other (clue 4). A's partner cannot be D (clue 6), B, C or F, so it is E: pairs {A, E} and {C, F}; D is the single at stop 3. Since D is above A, {A, E} is stop 2 and {C, F} is the top stop.
Floors: b < a < a + 2 < c ≤ 7 with a odd and b + a + (a + 2) + c = 17. a = 1 leaves no room for b; a = 5 makes D = 7 with no room for c. So a = 3, D = 5, and b + c = 9 with c ∈ {6, 7}, b ≤ 2 → b = 2, c = 7.
| Stop | Floor | Got off |
|---|---|---|
| 1 | 2 | B |
| 2 | 3 | A, E |
| 3 | 5 | D |
| 4 | 7 | C, F |
Arrival at floor 7 = 5 × 7 + 12 × 3 = 35 + 36 = 71 s. It halts 12 s and leaves at 83 s. Descending 7 floors takes 35 s, so it reaches the ground floor at 83 + 35 = 118 s.
Check: total = moving time (7 up + 7 down = 14 floors × 5 = 70 s) + 4 halts × 12 = 48 s → 118 s ✓.
The two pairs are {C, F} (clue 5) and A with one other (clue 4). A's partner cannot be D (clue 6), B, C or F, so it is E: pairs {A, E} and {C, F}; D is the single at stop 3. Since D is above A, {A, E} is stop 2 and {C, F} is the top stop.
Floors: b < a < a + 2 < c ≤ 7 with a odd and b + a + (a + 2) + c = 17. a = 1 leaves no room for b; a = 5 makes D = 7 with no room for c. So a = 3, D = 5, and b + c = 9 with c ∈ {6, 7}, b ≤ 2 → b = 2, c = 7.
| Stop | Floor | Got off |
|---|---|---|
| 1 | 2 | B |
| 2 | 3 | A, E |
| 3 | 5 | D |
| 4 | 7 | C, F |
Arrival at floor 7 = 5 × 7 + 12 × 3 = 35 + 36 = 71 s. It halts 12 s and leaves at 83 s. Descending 7 floors takes 35 s, so it reaches the ground floor at 83 + 35 = 118 s.
Check: total = moving time (7 up + 7 down = 14 floors × 5 = 70 s) + 4 halts × 12 = 48 s → 118 s ✓.
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